Divide the following complex numbers. $\dfrac{-1+5i}{1-i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${1+i}$. $ \dfrac{-1+5i}{1-i} = \dfrac{-1+5i}{1-i} \cdot \dfrac{{1+i}}{{1+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-1+5i) \cdot (1+i)} {1^2 - (-i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-1+5i) \cdot (1+i)} {(1)^2 - (-i)^2} $ $ = \dfrac{(-1+5i) \cdot (1+i)} {1 + 1} $ $ = \dfrac{(-1+5i) \cdot (1+i)} {2} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-1+5i}) \cdot ({1+i})} {2} $ $ = \dfrac{{-1} \cdot {1} + {5} \cdot {1 i} + {-1} \cdot {1 i} + {5} \cdot {1 i^2}} {2} $ $ = \dfrac{-1 + 5i - 1i + 5 i^2} {2} $ Finally, simplify the fraction. $ \dfrac{-1 + 5i - 1i - 5} {2} = \dfrac{-6 + 4i} {2} = -3+2i $